Grassmann Coordinates - PDF Free Download

Grassmann Coordinates and tableaux Matthew Junge Autumn 2012

Goals 1 Describe the classical embedding G(k, n) P N. 2 Characterize the image of the embedding quadratic relations. vanishing polynomials. 3 Reinterpret in terms of varieties and ideals. 4 Application: classify representations over GL n (C).

What is a Grassmannian? A Grassmannian G(k, n) is the set of all k-dimensional subspaces of C n. For example, G(1, 3) = P 2 where we identify all lines. G(k, n) can be given a topology by embedding it as a subspace of P N.

The Embedding Fix n, k and fix a basis for C n. Let S k G(k, n) be k-dimensional subspace. Goal: Map S k to a point in P (n k) 1.

S k p I P (n k) 1 Let α 1,..., α k C n be a basis for S k, and let A = the corresponding k n matrix. α 1. α k Let I = i 1... i k with each 1 i j n and i 1 < i 2 < < i k. Let A I denote the k k submatrix obtained by selecting the columns with suffixes i 1,..., i k. We define coordinate functions Φ I (A I ) = det A I := p I. This gives a map Φ : G(n, k) P (n k) 1 be S k (..., p I,...), I. }{{} ( k)-tuple n

Details About Embedding Proposition Φ is injective. Messy argument with coordinates.

Details About Embedding Proposition Φ is injective. Messy argument with coordinates. Proposition Φ is not surjective.

The Plücker Relations G(k, n) P (n k) 1. Goal: Characterize the image of G(k, n). Let X = Φ(G(k, n)). The points in X satisfy certain quadratic relations.

The Plücker Relations G(k, n) P (n k) 1. Goal: Characterize the image of G(k, n). Let X = Φ(G(k, n)). The points in X satisfy certain quadratic relations. Proposition The points in X do not satisfy any linear relations.

The Plücker Relations Theorem (Plücker Relations) Fix p X. For all 1 s n and any coordinates p I, p J with I = i 1... i k and J = j 1... j k it holds that p I p J = k p i1...i s 1 j λ i s+1...i k p j1 j λ 1 i sj λ+1...j k. λ=1 Theorem (Surjectivity Theorem) If p P N satisfies the Plücker relations then there is a k-space S k P n with coordinate p.

Basis Theorem I Definition Let 1 i 1,..., i k 1 n and let 1 j 1,..., j k+1 n be distinct numbers. Denote these two choices by I and J. We define a quadratic basis polynomial k+1 F IJ (P) = ( 1) λ P i1...i k 1 j λ P j1...j λ 1 j λ+1 j k+1 λ=1 with the P L inderminates.

Basis Theorem I Proposition For all p X and all I, J it holds that F IJ (p) = 0.

Basis Theorem I Proposition For all p X and all I, J it holds that F IJ (p) = 0. But what if G(p) = 0? For arbitrary homogeneuos G.

Basis Theorem Theorem (Basis Theorem I) If G(P) is a homogeneous polynomial in the indeterminates..., P L,... with L = l 1... l k such that G(p) = 0, p X then G(P) = I,J A IJ (P)F IJ (P), I = i 1... i k 1, J = j 1... j k+1 (1) with the F IJ quadratic basis polynomials and A IJ homogeneous polynomials in the P L.

Summary We can embed G(k, n) into P (n k) 1. The image consists of points satisfying certain quadratic (Plücker) relations. The set of polynomials which vanish on the image is generated by a set of quadratic polynomials. Up Next: This can all be reformulated and proven in terms of varieties and ideals in a coordinate free way.

Coordinate-Free Version Let E be a C-vector space, recall that ( d d ) E = E /T with T = {v 1 v d sign (σ)v σ(1) v σ(d) }. 1 1 2 d E is multilinear. d E is anticommutative.

Coordinate-Free Embedding ( d ) Fix S n d G(n d, n). Will map G(n d, n) P E via S n d H Sn d.

Coordinate-Free Embedding ( d ) Fix S n d G(n d, n). Will map G(n d, n) P E via S n d H Sn d. The kernel of the map d E d (E/S n d ) is a hyperplane H Sn d d E. Recall that P (E) is the quotient of E in which we identify all hyperplanes.

Polynomials on P ( d E) For any v 1,... v d E we can define a linear form on H P ( d E). d E π ( d E) /H := L For f L. ( ) d π E L (v 1 v d )(H) := (v 1 v d )(L ) (π f )(v 1 v d ) { 0 0 Products of the v 1 v d live in Sym ( ) d E.

Plücker Relations Theorem (The Plücker Relations/Surjectivity) The Plücker embedding is a bijection from G(n d, n) to the subvariety of P ( d E) defined by the quadratic equations (v 1 v d ) (w 1 w d ) = i 1 <i 2 < <i k (v 1 w 1 w k v d ) (v i1 v ik w k+1 w d )

The Basis Theorem II Theorem (The Basis Theorem II) Let Q be the ideal generated by the Plücker Relations. It holds that I(Z( Q)) = Q. Proof. We will prove that Q is prime. The Nullstellensatz immediately implies the result.

Proving Primality of Q. Short-Story: Goal is to show that Sym ( d E ) / Q is an integral domain. Will prove it embeds as a subring of a polynomial ring. Obtain a classification of polynomial representations over GL n (C).

Tableaux Let E be a C-module. For fixed n, we let λ denote a weakly decreasing partition of n, i.e. for n = 16 a partition λ could be λ = (6, 4, 4, 2) 6 + 4 + 4 + 2 = 16. The associated tableau (also denoted λ) is

Constructing the Schur Module: Step 1/4 From each λ we can construct a particular C-module E λ. Start with cartesian product E λ Instead of n-tuples - put elements in boxes.

Constructing the Schur Module: Step 1/4 From each λ we can construct a particular C-module E λ. Start with cartesian product E λ Instead of n-tuples - put elements in boxes. If n = 5 and λ = (2, 2, 1) we have an element v E λ is written v 1 v 4 v 2 v 5 v 3

Constructing the Schur Module: Step 2/4 Let λ have s columns and let d i, i = 1,..., s denote the length of the i th column. For example, E λ s d i E : v v i=1 1 v 1 v 4 v 2 v 5 v 3 (v1 v 2 v 3 ) (v 4 v 5 )

The Quadratic Relations: Step 3/4 Let Q λ be the submodule generated by v w The sum is over all w obtained from v with an exchange between two given columns with a given subset of boxes in the right chosen column. 1 6 1115 2 7 12 6 3 8 13 4 9 14 5 10 }{{} v 11 6 1 15 2 7 12 6 13 8 3 4 9 5 1410 }{{} w

The Schur Module: Step 4/4 s d i E λ := E /Q λ. i=1 1 1 λ = } {{ } n times then E λ = Sym n (E). 2 λ = then E λ = n E.

E λ in Coordinates Let e 1,..., e n be a basis for E. Fill λ with the e i. Weakly increasing across rows. Strictly increasing down columns. Each such arrangement, T, is called a standard filling. The image of this element in E λ will be denoted by e T. e 1 e 2 e 2 e 3 e 4 e 5 e 5 e 5 e T E λ

A New Polynomial Ring C[Z ] := C[..., Z i,j,...], i = 1,..., m j = 1,..., n For d m choose 0 i 1 i d n. Define the polynomial Z 1,i1 Z 1,id D i1...i d = det Z d,i1 Z d,id

A New Polynomial Ring C[Z ] := C[..., Z i,j,...], i = 1,..., m j = 1,..., n For d m choose 0 i 1 i d n. Define the polynomial Z 1,i1 Z 1,id D i1...i d = det Z d,i1 Z d,id For an arbitrary filling T of λ with the numbers {1,..., n}, Corollary D T = s i=1 D T (1,i),T (2,i),...,T (di,i), The map e T D T is an injective homomorphism E λ C[Z ] and its image D λ is free on the polynomials D T.

Tying it Together λ with s columns with lengths d i each occurring with multiplicity a i. E λ Sym a 1 ( d 1 E) Sym as ( d s E)/Q λ. Define S (E; d 1,..., d s ) := E λ, Q := Q λ. (a 1,...,a s) (a 1,...,a s) R := Sym a 1 ( d 1 E) Sym as ( d s E) (a 1,...,a s) R/Q = (a 1,...,a s) E λ

Putting it Together. We now have 1 R/Q = (a 1,...,a s) E λ. 2 E λ D λ C[Z ] under the map e T D T Proposition Q is a prime ideal. Proof. R/Q D λ C[Z ] via e T D T. D λ remains direct (requires proof) and thus is a subring. A subring of a polynomial ring is an integral domain. Q is prime.

What about Q? Back to G(n d, n) and Q. Corresponds to λ has columns of length d. E λ = ( d ) ( Sym a E /Q λ = Sym d ) E / Q. a a Which we just proved embeds as a subring of a polynomial ring. Hence Q is prime as a special case. Last item of business (time pending): Why is D λ direct?

Some Representation Theory A representation of GL(n, C) on C is a homomorphism V : GL(n, C) GL(m, C) for some m. Let X i,j : GL(n, C) C be the coordinate function with 1 i, j n. We say that a representation, V, is polynomial if there is a basis v 1,..., v m of V such that for g GL(n, C) we have gv b = a f ab (g)v a, 1 a, b n. With f ab C[X ij ] (i.e. f ab is a polynomial).

E λ as a Polynomial Representation Let λ = n, e T E λ acts on a matrix g GL(m, C) via the formula g e T = g i1,j 1 g im,j m e T where the sum is taken over the n m fillings of T of obtained from T by replacing the entries (j 1,..., j m ) by (i 1,..., i m ). Theorem As λ varies over all tableaux the E λ classify uniquely all irreducible polynomial representations of GL(n, C).

Conclusion 1 G(k, n) P N in coordinates and G(n d, n) P ( d E) via a coordinate free way. 2 Can classify the vanishing polynomials on the respective images. 3 All polynomial representations of GL(m, k) have the form E λ.